SARAJEVO JOURNAL OF MATHEMATICS
Vol.10 (22) (2014), 47–54
DOI: 10.5644/SJM.10.1.06
STARLIKENESS OF DOUBLE INTEGRAL OPERATORS
RASOUL AGHALARY AND SANTOSH JOSHI
Abstract. In this paper we investigate starlikeness of double integral
operators by using second-order differential inequalities. We shall give
some interesting conditions for f (z) defined by double integral operators
to be starlike of order β.
1. Introduction
Let H denote the class of analytic functions in the open unit disc ∆ =
{z ∈ C : |z| < 1} and An denote the class of all functions f in H such that
f has the form
∞
X
f (z) = z +
ak z k ,
k=n+1
where n ∈ N is fixed. Set A := A1 . For a positive integer n and a ∈ C, we
define the following class of analytic functions:
H[a, n] = {f ∈ H : f (z) = a + an z n + an+1 z n+1 + · · · , z ∈ ∆}.
Also let S ∗ (α) denote the familiar class of functions in A that are starlike
of order α. It is well known that this class is analytically characterized by
0 zf (z)
> α (z ∈ ∆, 0 ≤ α < 1).
Re
f (z)
For f and g in H, a function f is subordinate to g, written as f (z) ≺ g(z),
if there is an analytic function ω satisfying ω(0) = 0 and |ω(z)| < 1, such
that f (z) = g(ω(z), z ∈ ∆. If g is univalent in ∆, then f is subordinate to
g which is equivalent to f (∆) ⊆ g(∆) and f (0) = g(0).
Many authors have given sufficient conditions for starlikeness of analytic
functions (see [1], [2], [6], [7], [8]). In [4], Kuroki and Owa proved the
following results:
2010 Mathematics Subject Classification. Primary: 30C45; Secondary: 30C80.
Key words and phrases. Starlike functions, differential subordination, integral operators.
48
RASOUL AGHALARY AND SANTOSH JOSHI
Theorem 1.1. Let f (z) ∈ An and let 0 ≤ α < n and 0 ≤ β < 1. If f (z)
satisfies
(n + 1)(1 − β)(n − α)
|zf 00 (z) − α(f 0 (z) − 1)| <
n+1−β
then f (z) is starlike of order β in ∆.
Corollary 1.1. Let a function g(z) ∈ H satisfy
(n + 1)(1 − β)(n − α)
,
n+1−β
for some 0 ≤ α < n and 0 ≤ β < 1. Then the function f (z) given by
Z 1Z 1
n+1
g(rsz)rn−α−1 sn drds
f (z) = z + z
|g(z)| ≤
0
0
is starlike of order β in ∆.
We note that by setting β = 0 in Theorem 1.1 and Corollary 1.1 we obtain
the results of Miller and Mocanu [5].
However the case α = n and α ∈ C, which does produce a slightly different
implication, have not been discussed in [4] and [5]. In this paper we aim to
investigate such cases.
For establishing our results, we need the following lemma concerning differential subordinations.
Lemma 1.1. [3] Let h(z) be a convex function with h(0) = a and let Reγ >
0. If p(z) ∈ H[a, n] and
p(z) +
zp0 (z)
≺ h(z),
γ
then
p(z) ≺ q(z) ≺ h(z),
where
γ
q(z) = γ/n
nz
Z
z
h(t)tγ/n−1 dt.
0
This result is sharp.
2. Main results
Theorem 2.1. Let f (z) = z + an+1 z n+1 + · · · ∈ An be such that |an+1 | <
1/(n + 1). Also suppose that 0 < λ ≤ 1 − (n + 1)|an+1 |. If f (z) satisfies
|zf 00 (z) − n(f 0 (z) − 1)| < λ
(2.1)
then f (z) is starlike of order β in ∆, where
β=
(n + 2)[1 − (n + 1)|an+1 | − λ]
.
(n + 2) − (n + 2)|an+1 | − λ
(2.2)
STARLIKENESS OF DOUBLE INTEGRAL OPERATORS
49
Proof. We can rewrite the inequality (2.1) in terms of subordination as
zf 00 (z) − n(f 0 (z) − 1) ≺ λz.
(2.3)
If we set
p(z) := f 0 (z) − (n + 1)
f (z)
= −n + an+2 z n+1 + · · · ∈ H[−n, n + 1], (2.4)
z
then p(z) is analytic in ∆, p(0) = −n and p(k) (0) = 0 for k = 1, 2, . . . , n − 1.
Further, (2.3) is seen to be equivalent to
p(z) + zp0 (z) ≺ −n + λz.
(2.5)
Applying Lemma 1.1 to (2.5), we obtain
Z z
1
[−n + λt]t1/(n+1)−1 dt,
p(z) ≺
(n + 1)z 1/(n+1) 0
or equivalently
f (z)
λ
≺ −n +
z.
z
n+2
We can rewrite the relation (2.6) as
f 0 (z) − (n + 1)
f 0 (z) − (n + 1)
f (z)
λ
= −n +
ω(z),
z
n+2
(2.6)
(2.7)
for some ω ∈ Bn . Here,
Bn = {ω ∈ H : ω(0) = ω 0 (0) = · · · = ω (n) (0) = 0, and
|ω(z)| < 1 for z ∈ ∆}.
If we consider
f (z)
−1
z
then the relation (2.7) can be written as
g(z) =
zg 0 (z) − ng(z) =
λ
ω(z).
n+2
An algebraic computation implies that
λ
g(z) = an+1 z +
n+2
n
Z
0
1
ω(tz)
dt.
tn+1
As ω(z) ∈ Bn , Schwarz’s lemma gives that |ω(z)| ≤ |z|n+1 for z ∈ ∆ and
therefore,
λ
n
|g(z)| ≤ |z| |an+1 | +
|z| , z ∈ ∆
n+2
50
RASOUL AGHALARY AND SANTOSH JOSHI
which is
f (z)
< |an+1 | + λ ,
−
1
z
n+2
z ∈ ∆.
So that
f (z) λ
z > 1 − |an+1 | − n + 2 =: k.
Also from (2.6) it follows that
0
f (z) − (1 + n) f (z) < n + λ .
z n+2
Combining these last two inequalities, we see that
0
zf (z)
f (z) zf 0 (z)
k
− (1 + n) < − (1 + n)
f (z)
z
f (z)
0
λ
f (z) = f (z) − (1 + n)
<n+
z n+2
which simplifies to
0
zf (z)
n
λ
< +
−
(1
+
n)
f (z)
k (n + 2)k .
This gives us
Re
where
β=
zf 0 (z)
f (z)
> β,
(n + 2)[1 − (n + 1)|an+1 | − λ]
.
(n + 2) − (n + 2)|an+1 | − λ
We introduce the following examples to illustrate Theorem 2.1:
Example 2.1. Let the function f (z) = z + an+1 z n+1 + an+2 z n+2 be such
that (n + 1)|an+1 | + (n + 2)|an+2 | < 1. Then the function f is starlike.
By putting an+1 = 0 and λ = (n + 2)(1 − γ)/(n + 2 − γ), (0 ≤ γ < 1) in
Theorem 2.1 we obtain
Example 2.2. The function f (z) = z +
starlike of order γ.
(n+2)(1−γ) n+2
, (0
(n+2−γ) z
≤ γ < 1) is
Remark 2.1. We note that by using Theorem 2.1 we proved that Example
2.1 holds true, while from Theorem 1.1 it is not easy to show.
By making use of Theorem 2.1, we obtain the following result concerning
the starlikeness of the double integral operator.
STARLIKENESS OF DOUBLE INTEGRAL OPERATORS
51
Theorem 2.2. Let c be a complex number with |c| < 1/(n + 1) and 0 < λ ≤
1 − |c|(n + 1). Also suppose that g(z) ∈ H satisfies
|g(z)| < λ,
then the function f (z) given by
f (z) = z + cz
n+1
+z
n+2
1Z 1
Z
0
g(rsz)sn+1 drds
0
is starlike of order β in ∆, where β is given by (2.2).
Proof. First let the function f (z) ∈ An be the solution of the differential
equation
zf 00 (z) − n(f 0 (z) − 1) = z n+1 g(z),
(2.8)
then we have
|zf 00 (z) − n(f 0 (z) − 1)| ≤ |z n+1 ||g(z)| < λ.
Thus, from Theorem 2.1, we conclude that f (z) is starlike of order β, where
β is given by (2.2). The solution of (2.8) can be obtained by integrating
twice. By setting ϕ(z) = f 0 (z) − 1, we can rewrite the equation (2.8) as
zϕ0 (z) − ϕ(z) = z n+1 g(z),
which has solution ϕ(z) given by
n
ϕ(z) = cz + z
n+1
1
Z
g(rz)dr.
0
Since ϕ(z) = f 0 (z) − 1, we have
0
n
f (z) − 1 = cz + z
n+1
Z
1
g(rz)dr,
0
that is
f (z) = z + cz n+1 + z n+2
Z
0
1Z 1
g(rsz)sn+1 drds.
0
Example 2.3. Let c and k be complex numbers such that (n+1)|c|+|k| < 1.
Then for the function g(z) = kz, we have
f (z) = z + cz n+1 +
and this function is starlike.
k
z n+3
2(n + 3)
52
RASOUL AGHALARY AND SANTOSH JOSHI
Example 2.4. Let c and k be complex numbers such that (n+1)|c|+|k| < 1.
Then for the function g(z) = k, we have
f (z) = z + cz n+1 +
k
z n+2
(n + 2)
and this function is starlike.
In the next results we generalize the results obtained in [4].
Theorem 2.3. Let 0 ≤ β < 1 and α be a complex number with max{0, |α| +
β − 1} ≤ Reα < n and let f (z) ∈ An . If f (z) satisfies
|zf 00 (z) − α(f 0 (z) − 1)| <
(n − Reα)(n + 1)[1 + Reα − |α| − β]
,
n+1−β
(2.9)
then f (z) is starlike of order β in ∆.
Proof. The inequality (2.9) can be written as follows
zf 00 (z) − α(f 0 (z) − 1) ≺
(n − Reα)(n + 1)[1 + Reα − |α| − β]
z.
n+1−β
(2.10)
If we set
p(z) = f 0 (z) − (1 + α)f (z)/z
= −α + (n − α)an+1 z n+1 + ... ∈ H[−α, n],
then (2.10) becomes
p(z) + zp0 (z) ≺ −α +
(n − Reα)(n + 1)[1 + Reα − |α| − β]
z.
n+1−β
Applying Lemma 1.1 as well as using the proof of Theorem 2.1, we conclude
that
Z z
1
(n − Reα)(n + 1)[1 + Reα − |α| − β]
p(z) ≺ 1/n
−α +
t t1/n−1 dt,
n+1−β
nz
0
or equivalently
p(z) ≺ −α +
(n − Reα)[1 + Reα − |α| − β]
z.
n+1−β
(2.11)
Hence from (2.11) it follows that
f 0 (z) − (1 + α)
f (z)
(n − Reα)[1 + Reα − |α| − β]
≺ −α +
z.
z
n+1−β
(2.12)
We can rewrite (2.12) as
f 0 (z) − (1 + α)
f (z)
(n − Reα)[1 + Reα − |α| − β]
= −α +
ω(z),
z
n+1−β
(2.13)
STARLIKENESS OF DOUBLE INTEGRAL OPERATORS
53
for some ω(z) ∈ Bn−1 . If we consider
g(z) =
f (z)
− 1,
z
then (2.13) can be rewritten as
zg 0 (z) − αg(z) =
(n − Reα)[1 + Reα − |α| − β]
ω(z).
n+1−β
An algebraic computation implies that
(n − Reα)[1 + Reα − |α| − β]
g(z) =
n+1−β
Z
0
1
ω(tz)
dt.
tα+1
As ω(z) ∈ Bn−1 , Schwarz’s lemma gives that |ω(z)| < |z|n for z ∈ ∆ and
therefore
1 + Reα − |α| − β n
|g(z)| <
|z| ,
n+1−β
which is
1 + Reα − |α| − β
f (z)
,
z − 1 <
n+1−β
or
f (z) n − Reα + |α|
1 + Reα − |α| − β
=
.
z >1−
n+1−β
n+1−β
Also from (2.12) we obtain
0
f (z) − (1 + α) f (z) < |α| + (n − Reα)[1 + Reα − |α| − β]
z n+1−β
=
(1 + Reα − β)(|α| + n − Reα)
.
n+1−β
Combining these two last results, we observe that
f (z) zf 0 (z)
n − Reα + |α| zf 0 (z)
− (1 + α) < − (1 + α)
n+1−β
f (z)
z
f (z)
f (z) (1 + Reα − β)(|α| + n − Reα)
,
= zf 0 (z) − (1 + α)
<
z n+1−β
which implies that
0
zf (z)
f (z) − (1 + α) < 1 + Reα − β,
or
zf 0 (z)
> (1 + Reα) − (1 + Reα − β) = β.
f (z)
and the proof is complete.
Re
54
RASOUL AGHALARY AND SANTOSH JOSHI
We note that when in Theorem 2.3, α is a real number we get Theorem
1.1. In the same way is in the proof of Theorem 2.2 we obtain the following
result, the details are thus omitted.
Theorem 2.4. Let a function g(z) ∈ H satisfy
(n − Reα)(n + 1)[1 + Reα − |α| − β]
,
n+1−β
for some 0 ≤ β < 1 and max{0, |α| + β − 1} ≤ Reα < n. Then the function
f (z) given by
Z 1Z 1
n+1
g(rsz)rn−α−1 sn drds,
f (z) = z + z
|g(z)| ≤
0
0
is starlike of order β in ∆.
References
[1] R. Aghalary and A. Ebadian, New coefficient conditions for functions starlike with
respect to symmetric points, Serdica Math. J., 36 (2010), 211–222.
[2] R. M. Ali, S. K. Lee, K. G. Subramanian and A. Swaminathan, A third-order differential equation and starlikeness of a double integral operator, Abstr. Appl. Anal., (2011),
Art. ID 901235, 10 pp.
[3] D. J. Hallenbeck and St. Ruscheweyh, Subordination by convex functions, Proc. Amer.
Math. Soc., 52 (1975), 191–195.
[4] K. Kuroki and S. Owa, Double integral operators concerning starlike of order β, Int.
J. Differ. Equ., (2009) 1–13.
[5] S. S. Miller and P. T. Mocanu, Double integral starlike operators, Integral Transforms
Spec. Funct., 19 (7-8) (2008), 591–597.
[6] M. Obradović, Simple sufficient conditions for univalence, Mat. Vesnik, 49 (1997),
241–244.
[7] J. Sokól, On sufficient condition for starlikeness of certain integral of analytic function,
J. Math, Appl., 26 (2006), 127–130.
[8] J. Sokól, On a condition for α-starlikeness, J. Math. Anal. Appl., 352 (2009), 696–701.
(Received: May 30, 2013)
Rasoul Aghalary
Department of Mathematics
Faculty of Science
Urmia University
Urmia, Iran
[email protected]
Santosh Joshi
Department of Mathematics
Walechand college of Engineering
Sangli, India
[email protected]